Question

Find the equation of the tangent and the normal to the following curve at the indicated point  \[x^\frac{2}{3} + y^\frac{2}{3}\] = 2 at (1, 1) ?

Answer 1

\[x^\frac{2}{3} + y^\frac{2}{3} = 2\]

\[\text { Differentiating both sides w.r.t.x}, \]

\[\frac{2}{3} x^\frac{- 1}{3} + \frac{2}{3} y^\frac{- 1}{3} \frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- x^\frac{- 1}{3}}{y^\frac{- 1}{3}} = \frac{- y^\frac{1}{3}}{x^\frac{1}{3}}\]

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) =\frac{- 1}{1}=-1\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( 1, 1 \right)\]

\[\text { Equation of tangent is,}\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - 1 = - 1\left( x - 1 \right)\]

\[ \Rightarrow y - 1 = - x + 1\]

\[ \Rightarrow x + y - 2 = 0\]

\[\text { Equation of normal is },\]

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - 1 = 1\left( x - 1 \right)\]

\[ \Rightarrow y - 1 = x - 1\]

\[ \Rightarrow y - x = 0\]